put 4x^2-y^2-8x-4y+16=0 in standard form what are the vertices and centre of the curve

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The equation 4x^2-y^2-8x-4y+16=0 is that of a hyperbola.

4x^2-y^2-8x-4y+16=0

=> 4x^2 - 8x + 4 - (y^2 + 4y + 4) = 4 - 4 - 16

=> 4(x - 1)^2 - (y + 2)^2 = -16

=> `(x - 1)^2/(-4) - (y + 2)^2/(-16) = 1`

=> `(y + 2)^2/(4^2) - (x - 1)^2/(2^2) = 1`

The center of the hyperbola is (1, -2) and the vertices are (1, -4) and (1, 0)

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